Dampmaskin

Battery drain

Often people seems to be somewhat confused about how battery drain is calculated, so here is a short summary.

But first: There are two main types of mods: Regulated, and unregulated (mechanical). Putting a Kick in a mech mod automagically makes it a regulated mod.

We will start with the simplest type, and then use a little more time on the more complex type.

Unregulated mods

Unregulated (mech) mods are very simple. They have one circuit. This circuit can be easily modelled using Ohms law. In other words, if you have a basic understanding of Ohm's law, that is all it takes.

The battery drain is determined by two factors:

The resistance of the atomizer, which is more or less constant after you have built your coil(s).

The voltage of the battery, which gets lower as the battery drains. Using a single battery, it starts at 4.2V or 4.1V, and sinks a volt or less (measured under load) before the battery needs to be recharged.

There is not much more to mech mods. The circuit is about as simple as a circuit can get. The voltage hitting the atomizer is the voltage from the battery (minus the tiny voltage lost in the switch and conductors). The current flowing through the battery is the current flowing through the atomizer. A multimeter and some straightforward use of Ohm's law you will give you all the numbers you need.

Regulated mods

Regulated mods are not as simple as mechs, and not so easy to model. But even though they are much more complex than mechs, with some selective simplification we can ignore most of the complexity.

So let us break these mods down into two circuits and a black box. This makes our regulated mod model little more than twice as complex as our mech mod model.

The two main circuits of a regulated mod, is the battery side, and the atomizer side, and never the twain shall meet. The regulator circuit takes care of that. The regulator can have lots of more or less advanced circuits in itself, and they all use a little bit of power, but for the most part we can envision it as a black box separating the battery circuit from the atty circuit.

On the atomizer side, the voltage is (ideally) whatever the user has selected, but take note that some APVs promise more than they keep: Unless you know that your APV is accurate, you should probably measure and confirm the output voltage.

Variable wattage devices simply measure the atomizer resistance, and uses this measurement and Ohm's law to calculate what voltage to set.

When you know the voltage and resistance, you can calculate the current and power.

On the battery side, the voltage is whatever the battery has in it at the moment (just like with mech mods). The power is whatever the regulator needs to pull in order to produce the desired voltage in the atomizer. And the current is the power divided by the ever-decreasing battery voltage.

Since we are looking at two separate circuits, you can never mix numbers from both sides of the regulator in your calculations. For instance: You cannot determine the current drained from the battery by measuring the resistance of the coil and the voltage of the battery. Using the current from one circuit, and the voltage from a different circuit, will result in a nonsensical answer. Nor can you determine the current going through the coil by determining the current from the battery. They are two different circuits. The regulator makes sure of that.

But not all hope is lost. There is one way to "transfer" information from one side of the regulator to the other. The trick is simple: Use the power (wattage).

The power hitting the atomizer equals the power flowing from the battery, minus the power lost in (or used by) the regulator circuit.

The efficiency of these regulator circuits are typically between 80-95%. In practice this means that the regulator "steals" about a tenth of the power between the battery and the atomizer.

Knowing this, we can use our multimeter and Ohm's law to calculate what is going on at either side of the regulator. Then we can convert it to watts, and voila! Subtracting (or adding) the loss in the regulator circuit, we now know the watts on the other side as well.

So we use Ohm's law again, break down the power to current and voltage, and that's that: Now we have all the numbers we need.

And that's it.

As you can see, there is no voodoo going on in regulated mods. And if there is, we just tuck it away in a black box and assign an efficiency percentage to it.

If you can use Ohms law once, you can use it twice. So now there is really nothing stopping you from figuring out what is going on in your mod, whether it's one of our future robotic overlords that you're vaping on, or a simple metal tube with a battery in it. Vape on, and vape safely!

Update: After I wrote this blog post, it occured to me that I could easily model this, so I made another calculator for you.

Comments

Nice write up and very interesting. But I have some questions. I'm new to vaping and I don't know anything about electric and I've been away from college math for close to 3 decades so you can tell me I don't know what the f-ck I'm talking about because I don't. Anyway...

Some thoughts and questions as I read this:

In a mechanical, does wattage decrease as the battery drains? Seems like it should - less battery voltage, less watts. Is that right? Which is the reason for a kick or Regulated mods. So you can maintain the same wattage even though the battery voltage is draining as the mod's being fired.

In my very short experience with a mechanical though, it seems like I get the same hit (power=wattage?) while the battery drains until it should be recharged. And if the wattage declines as the battery drains, does the wattage go down in a line or a curve? A curve like it slowly loses wattage then a quick drop when the battery voltage falls below a certain point like 3.6 V? I haven't yet run my battery so low on my mechanical that it stopped working.

And in a Regulated (vv/vw), the wattage stays exactly the same (by regulation) then when the voltage hits a certain low point, that's it, there's not enough voltage to main the wattage that was set so, nothing, the mod won't fire. Time to recharge the battery. But if this is true, and that point was hit, could you turn down the wattage setting and get the regulated mod to continue to still fire for a short time? Because the lower volts would still allow a lower wattage?

In a mechanical, does wattage decrease as the battery drains? Seems like it should - less battery voltage, less watts. Is that right?

Yes, it does, as per Ohm's/Joule's law.

Originally Posted by devilchasnme

In my very short experience with a mechanical though, it seems like I get the same hit (power=wattage?) while the battery drains until it should be recharged.

Sounds like your coil is very forgiving. You are not getting the same power, but as long as you're getting a similar vape, all is good, right?

Originally Posted by devilchasnme

And if the wattage declines as the battery drains, does the wattage go down in a line or a curve? A curve like it slowly loses wattage then a quick drop when the battery voltage falls below a certain point like 3.6 V? I haven't yet run my battery so low on my mechanical that it stopped working.

This is determined by two things: The relationship between voltage and power, and the characteristics of the battery (how voltage drops as the battery is drained).

I am not an expert on batteries, but I assume the voltage drop is far from linear. Other than that I have to admit that at the moment of writing this, I have less than 24 hours of personal experience with mech mods, (vaping on my brand new nemmy clone right now, though ) so I simply don't know.

When it comes to the relationship between voltage and power, I think that for a constant resistance, the relationship is parabolic. See Wolfram Alpha for the plot.

However, I suspect that the battery characteristics is the most important factor here, and I'm sorry to say I cannot help you with that.

Originally Posted by devilchasnme

And in a Regulated (vv/vw), the wattage stays exactly the same (by regulation) then when the voltage hits a certain low point, that's it, there's not enough voltage to main the wattage that was set so, nothing, the mod won't fire. Time to recharge the battery. But if this is true, and that point was hit, could you turn down the wattage setting and get the regulated mod to continue to still fire for a short time? Because the lower volts would still allow a lower wattage?

Whether the electronics can keep up or not I haven't given much thought, but it is not very relevant, because the main reason for regulated mods cutting out, is preserving the battery. Li-ion batteries fail permanently if the voltage drops beneath a certain point. Approaching too closely to this limit it will also decrease battery health. Maybe you can "fool" an APV to fire a little bit more at a lower wattage, and it may or may not be detrimental to the battery, but I see no reason to push it.

Also note that many mods will fire at a lower than set voltage/power if the battery voltage or atomizer resistance is lower than the regulating circuit can handle. In my experience, setting my iTaste VV to 5 volts most often produces somewhere between 3.5 and 4.5 volts, depending on the resistance.

Hey Dampmaskin, just wanted to double check and clarify, if you're running at dual 18650 regulated mod, to figure out the battery drain, do you just change the voltage to 8.4, and is that calculating in series? I plan on getting the ipv v3, and just want to know if the calculations are correct and safe. According to ohms law, if I had two 20 amp batteries at continuous, in a series that would be 8.4/20=0.42 ohms, so that would be the lowest safe resistance I could go to with those batteries on a series box mod correct? and then 8.4820=168 watts, which calculates to how much power I can give it to hit 20 amps correct? Well I went ahead and used your calculator, and of course there is efficiency involved, does that account for voltage drop as well? With regulated mods, it uses more amps as the battery drains to hit that wattage..so setting it at 168 watts, actually uses 186.67 W on the battery, so I went ahead and lowered the wattage on the device part, all the way down to 150 watts which gave me 1 percent head room on the battery, but that's at a full charge at 8.4 with two batteries, so I lowered it down to a pretty much drained battery at 7.2 which made the amps go up, so then I lowered the wattage even more to 129 watts which gave me 0% headroom right below 20 amps, so basically I was wondering if I'm using your calculator correctly and if so I could basically run two 20 amp batteries at a .42 ohm build at about 120 watts safely? I am new to this and hope I understood it correctly and also was wondering how accurate this was? And is efficiency the same as voltage drop, or do you have to add in voltage drop as well to see how many watts you are truly getting?